使用sapply()在两个参数上循环uniroot() [英] Looping uniroot() on two arguments with sapply()

问题描述

此功能根据给定的直径，流量和长度来计算管道中的压力损失.

This function that calculates the pressure loss in a pipe given its diameter, flowrate, and length.

hazwil2 <- function(diam, flow, leng){
  psi2=((1/(2.31*100))*1050*((flow/140)^1.852)*leng*diam^-4.87)
  return(psi2)
}

我正在寻找将压力损失保持在2 psi之内的最小直径.直径范围在2到12英寸之间.使用uniroot()和两个任意值作为流和长度:

I'm looking for the smallest diameter that will keep the pressure loss within 2 psi. The range of diameters is between 2 and 12 inches. Using uniroot() and two arbitrary values for flow and length:

intercept   = 2L
uniroot(
  function(x) hazwil2(x, 100, 400) - intercept ,
  interval = c(2, 12)
)$root

现在我正在尝试在流和长度值的这个数据集上循环uniroot

Now I'm trying to loop uniroot on this dataset of values of flow and length

data <- read.csv(
  text = 
  "leng,flow
  100, 100
  200, 100
  300, 100
  100, 150
  200, 150
  300, 150
  100, 200
  200, 200
  300, 200",
  stringsAsFactors = FALSE
)

推荐答案

考虑对 f 和 l 参数进行 unitroot 调用的泛化，以及将其传递给mapply( m 适用)以逐元素向下迭代数据帧的等长向量(即列， flow 和 length ) ):

Consider generalizing your unitroot call for f and l params, and pass it in mapply (multiple apply) to iterate elementwise down the equal length vectors (i.e., columns, flow and length of dataframe):

find_root <- function(f, l) {
  intercept <- 2L
  uniroot(
    function(x) hazwil2(x, f, l) - intercept ,
    interval = c(2, 12)
  )$root
}

data$root <- mapply(find_root, data$flow, data$leng)

data
#   leng flow     root
# 1  100  100 2.681145
# 2  200  100 3.091243
# 3  300  100 3.359606
# 4  100  150 3.128141
# 5  200  150 3.606602
# 6  300  150 3.919727
# 7  100  200 3.489780
# 8  200  200 4.023564
# 9  300  200 4.372916

对于可能产生根本问题的较大集合，请考虑在tryCatch中包装函数调用以返回NA:

For a larger set which can potentially yield root issues, consider wrapping function call in tryCatch to return NA:

data <- expand.grid(leng = seq(100, 1000, by=100), flow = seq(10, 200, by=10))
data$root <- mapply(function(l,f) tryCatch(find_root(l,f), error=function(e) NA), 
                    data$flow, data$leng)

输出

head(data, 20)

#    leng flow     root
# 1   100   10       NA
# 2   200   10       NA
# 3   300   10       NA
# 4   400   10       NA
# 5   500   10       NA
# 6   600   10       NA
# 7   700   10       NA
# 8   800   10       NA
# 9   900   10       NA
# 10 1000   10       NA
# 11  100   20       NA
# 12  200   20       NA
# 13  300   20       NA
# 14  400   20       NA
# 15  500   20 2.023187
# 16  600   20 2.100367
# 17  700   20 2.167915
# 18  800   20 2.228178
# 19  900   20 2.282705
# 20 1000   20 2.332653

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